Given \(\vec{a}=\hat{i}+2\hat{j}+3\hat{k}\) and \(\vec{b}=2\hat{i}+3\hat{j}-5\hat{k}\)
\(\therefore \left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 2 & 3 \\ 2 & 3 & -5 \end{matrix} \right| =\hat { i } (-10-9)-\hat { j(-5-6)+\hat { k } (3-4)=-19\hat { i } +11\hat { j } -\hat { k } } \)
Now, \(\vec { a } .(\vec { a } \times \vec { b } )=(\hat { i } +2\hat { j } +3\hat { k } ).(-19\hat { i } +11\hat { j } -\hat { k } )=-19\hat { i } +11\hat { j } -\hat { k } \)
=-19+22-3=-22+22=0
This shows \(\vec{a}\) and \(\vec{a} \times \vec{b}\) are perpendicular to each other.

Given \(\overrightarrow{a}=\hat{i}-2\hat{j}+\hat{k}\)
\(\overrightarrow{b}=3\hat{i}-4\hat{j}-2\hat{k}\)|
\(\overrightarrow{a} . \overrightarrow{b}=(\hat{i}-2\hat{j}+\hat{k}).(3\hat{i}-4\hat{j}-2\hat{k})\)
= 1(3) - 2(-4) + 1(-2)
= 3 + 8 - 2 = 9
\(\therefore \overrightarrow{a} . \overrightarrow{b}=9\)

Let \(l={2\over \sqrt{2}},m={1\over2}\) and \(n={1\over2}\)
\(\therefore l^2+m^2+n^2=({1\over \sqrt{2}})^2+({1\over2})^2+({1\over2})^2\)\(={1\over2}+{1\over4}+{1\over 4}={2+1+1\over4}={4\over4}=1\)
Hence, the given ratios are direction cosines of some vector.

The condition is cos²\(\alpha\) + cos²\(\beta\) + cos²\(\gamma\) = 1
Here \(\alpha =30^o,\beta =45^o,\gamma=60^o\)
cos²\(\alpha\) + cos²\(\beta\) + cos²\(\gamma\) \(={3\over 4}+{1\over2}+{1\over4}\neq1.\)
There fore they are not direction angles of any vector.

80km, 60° south of west

The vector \(\overrightarrow{OQ}\) represents a displacement of 80 km, 60° south of west.

(i)

(ii) 

Let the given points be A(2, - 1, 3), B(4, 3, 1) and C(3, 1, 2).
Then \(\overrightarrow{OA}=2\hat{i}-\hat{j}+3\hat{k},\overrightarrow{OB}=4\hat{i}+3\hat{j}+\hat{k}\) and \(\overrightarrow{OC}=3\hat{i}+\hat{j}+2\hat{k}\)
Now,  \(\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=(4\hat{i}+3\hat{j}+\hat{k})-(2\hat{i}-\hat{j}+3\hat{k})=2\hat{i}+4\hat{j}-2\hat{k}\)
\(\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=(3\hat{i}+\hat{j}+2\hat{k})-(4\hat{i}+3\hat{j}+\hat{k})=-\hat{i}-2\hat{j}+\hat{k}\)
\(\overrightarrow{CA}=\overrightarrow{OA}-\overrightarrow{OC}=(2\hat{i}-\hat{j}+3\hat{k})-(3\hat{i}+\hat{j}+2\hat{k})=-\hat{i}-2\hat{j}+\hat{k}\)
Now,  \(\overrightarrow{AB}=2\hat{i}+4\hat{j}-2\hat{k}=-2(-\hat{i}-2\hat{j}+\hat{k})=-2\overrightarrow{BC}\)
\(\lambda\) = -2.
Thus\(\overrightarrow{AB}||\overrightarrow{BC}\) and B is a common points. 
Hence, the given points are collinear.

Let the position vector of the vertices of the quadrilateral ABCD be \(\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\)and\(\overrightarrow{d}\) respectively.
\(\therefore \overrightarrow{OA}=\overrightarrow{a}, \overrightarrow{OB}=\overrightarrow{b},\overrightarrow{OC}=\overrightarrow{c}\) and \(\overrightarrow{OD}=\overrightarrow{d}.\)
Since E and F are the mid-points of AC and BD respectively, we have
\(\overrightarrow{OE}={\overrightarrow{a}+\overrightarrow{c}\over 2}\) and \(\overrightarrow{OF}={\overrightarrow{b}+\overrightarrow{d}\over 2}\)
To prove that \(\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{CB}+\overrightarrow{CD}=4\overrightarrow{EF}\)
\(LHS=\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{CB}+\overrightarrow{CD}\)
\(=\overrightarrow{OB}-\overrightarrow{OA}+\overrightarrow{OD}-\overrightarrow{OA}+\overrightarrow{OB}-\overrightarrow{OC}+\overrightarrow{OD}-\overrightarrow{OC}\)
\(=\overrightarrow{b}-\overrightarrow{a}+\overrightarrow{d}-\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c}+\overrightarrow{d}-\overrightarrow{c}\)

\(=-2\overrightarrow{a}+2\overrightarrow{b}-2\overrightarrow{c}+2\overrightarrow{d}\)
\(=2[(\overrightarrow{b}+\overrightarrow{d})-(\overrightarrow{a}+\overrightarrow{c})]\)
\(=2[2\overrightarrow{OF}+2\overrightarrow{OE})\) [From (1)]
\(=4[\overrightarrow{OF}-\overrightarrow{OE}]=4.\overrightarrow{EF}=RHS\)
Hence proved.

Let the position vector of the vertices of the triangle be \(\overrightarrow{a},\overrightarrow{b}\)and \(\overrightarrow{c}\) respectively
\(\therefore \overrightarrow{OA}=\overrightarrow{a},\overrightarrow{OB}=\overrightarrow{b},\overrightarrow{OC}=\overrightarrow{c}.\)
Since D is the mid-point of BC,
\(\overrightarrow {OD}={\overrightarrow{b}+\overrightarrow{c}\over 2}\).....(1)
To prove that \(\overrightarrow{AB}+\overrightarrow{AC}=2\overrightarrow{AD}\)
LHS =\(\overrightarrow{AB}+\overrightarrow{AC}\)
=\(\overrightarrow{OB}-\overrightarrow{OA}+\overrightarrow{OC}-\overrightarrow{OA}\)
\(=\overrightarrow{b}-\overrightarrow{a}+\overrightarrow{c}-\overrightarrow{a}=\overrightarrow{b}+\overrightarrow{c}-2\overrightarrow{a}\)
RHS \(=2 \overrightarrow{AD}\)

=2\((\overrightarrow{OD}-\overrightarrow{OA})\)
\(=2({\overrightarrow{b}+\overrightarrow{c}\over 2}-\overrightarrow{a})\)       [From (1)]
\(=2({\overrightarrow{b}+\overrightarrow{c}-2\overrightarrow{a}\over 2})=\overrightarrow{b}+\overrightarrow{c}-2\overrightarrow{a}\)
\(\therefore LHS=RHS\)
Hence proved.

Given  \(\overrightarrow{PO}\) +\(\overrightarrow{OQ}\) = \(\overrightarrow{QO}\) +\(\overrightarrow{OR}\)
\(\Rightarrow \overrightarrow{PQ}=\overrightarrow{QR}\) [By triangle law of addition]
\(\Rightarrow \overrightarrow{PQ}=\overrightarrow{QR}\) and Q is a common point.
Hence, the points P, Q, R are collinear.

Let \(\overrightarrow{a}\) and \(\overrightarrow{b}\) be the position vectors of the points A and B.
\(\Rightarrow \overrightarrow{OA}=\overrightarrow{a}\) and \( \overrightarrow{OB}=\overrightarrow{b}\).
Let P divides the line segment AB in the ratio 1:2 and Q divides the line segment AB in the ratio 2 : 1
\(\therefore \overrightarrow{OP}={1.(\overrightarrow{OB})+2(\overrightarrow{OA})\over 1+2}={1(\overrightarrow{b})+2(\overrightarrow{a})\over 3}={\overrightarrow{b}+2\overrightarrow{a}\over 3}\)
and \( \overrightarrow{OQ}={2(\overrightarrow{OB})+1(\overrightarrow{OA})\over 2+1}={2\overrightarrow{b}+\overrightarrow{a}\over 3}={\overrightarrow{a}+2\overrightarrow{b}\over 3}\)
Hence, the required position vectors are \({\overrightarrow{b}+2\overrightarrow{a}\over 3}\)and \({\overrightarrow{a}+2\overrightarrow{b}\over 3}\).

\((\overrightarrow { a } +\overrightarrow { b } ).(\overrightarrow { a } -\overrightarrow { b } )=0\)
\(\Rightarrow \quad \overrightarrow { a } .\overrightarrow { b } -\overrightarrow { a } .\overrightarrow { b } +\overrightarrow { b } .\overrightarrow { a } -\overrightarrow { b } .\overrightarrow { b } =0\)
 \(\\ \left[ \because \overrightarrow { a } .\overrightarrow { b } =\overrightarrow { b } .\overrightarrow { a } \right] \)   
\(\Rightarrow \quad { |\overrightarrow { a } | }^{ 2 }-{ |\overrightarrow { b } | }^{ 2 }=0\)
\(\Rightarrow \quad { |\overrightarrow { a } | }^{ 2 }={ |\overrightarrow { b } | }^{ 2 }\)
\(\Rightarrow \quad |{ \overrightarrow { a } | }=|\overrightarrow { b } |\)

LHS =\(\overrightarrow{a}\times (\overrightarrow{b}+\overrightarrow{c})+\overrightarrow{b}\times (\overrightarrow{c}+\overrightarrow{a})+\overrightarrow{c}\times (\overrightarrow{a}+\overrightarrow{b})\)
=\(\vec { a } \times \vec { b } +\vec { a } \times \vec { c } +\vec { b } \times \vec { c } +\vec { b } \times \vec { a } +\vec { c } \times \vec { a } +\vec { c } \times \vec { b } \) (By associative property)   
\(\left[ \therefore \vec { b } \times \vec { a } =-\vec { a } \times \vec { b } \vec { c } \times \vec { a } =-\vec { a } \times \vec { c } \vec { c } \times \vec { b } =-\vec { b } \times \vec { c } \right] \)
= \(\vec { a } \times \vec { b } +\vec { a } \times \vec { c } -\vec { b } \times \vec { c } -\vec { b } \times \vec { a } -\vec { c } \times \vec { a } -\vec { c } \times \vec { b } \) = \(\vec{0}\) = RHS
Hence proved.

Let \(\overrightarrow{a}=5\hat{i}+3\hat{j}+4\hat{k},\) and \(\overrightarrow{b}=6\hat{i}-8\hat{j}-\hat{k}\) .
Let \(\theta\) be the angle between them.
\(cos \theta ={\overrightarrow{a}.\overrightarrow{b}\over|\overrightarrow{a}||\overrightarrow{b}|}={30-24-4\over \sqrt{50}\sqrt{101}}={\sqrt{2}\over 5\sqrt{101}}\Rightarrow \theta =cos^{-1}\left[ \sqrt{2}\over 5\sqrt{101} \right] \).

\(\vec{a} \perp \vec{b} \Rightarrow \quad \vec{a} \cdot \vec{b} =0 \)
\((\sin \theta \hat{i}+\cos \theta \hat{j}) \cdot(\hat{i}-\sqrt{3} \hat{j}+2 \hat{k}) =0 \)
\(\sin \theta-\sqrt{3} \cos \theta =0 \)
\(\sin \theta =\sqrt{3} \cos \theta \)
\(\frac{\sin \theta}{\cos \theta} =\sqrt{3} \)
\(\tan \theta =\sqrt{3} \)
\(\theta =60^{\circ} \text { or } \frac{\pi}{3} \)

\(\text { In } \Delta \mathrm{BCD}, \overrightarrow{\mathrm{BD}}=\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CD}}\)
\(=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}\)

\(\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1 \)
\(\cos ^{2} 60^{\circ}+\cos ^{2} 45^{\circ}+\cos ^{2} \gamma=1 \)
\(\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}+\cos ^{2} \gamma=1 \Rightarrow \cos ^{2} \gamma=1-\frac{1}{4}-\frac{1}{2} \)
\(\cos ^{2} \gamma=\frac{4-1-2}{4}=\frac{1}{4} \Rightarrow \cos \gamma=\frac{1}{2} \Rightarrow \gamma=60^{\circ} \)
\(\text { Angle between } \overrightarrow{O P} \text { and } z \text { axis is } 60^{\circ}\)

\(\underbrace{\overrightarrow{A B}}+ \overrightarrow{B C}+\overrightarrow{C D}+\overrightarrow{D A} \)
\(=\underbrace{\overrightarrow{A C}+\overrightarrow{C D}}+\overrightarrow{D A} \)
\(=\underbrace{\overrightarrow{A D}+\overrightarrow{D A}} \)
\(=\overrightarrow{A A}=\overrightarrow{0} . \)